Simplify the following expression and state the condition under which the simplification is valid. $z = \dfrac{-6r^2 + 6r + 252}{8r^2 + 112r + 384}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ z = \dfrac {-6(r^2 - r - 42)} {8(r^2 + 14r + 48)} $ $ z = -\dfrac{6}{8} \cdot \dfrac{r^2 - r - 42}{r^2 + 14r + 48} $ Simplify: $ z = - \dfrac{3}{4} \cdot \dfrac{r^2 - r - 42}{r^2 + 14r + 48}$ Next factor the numerator and denominator. $ z = - \dfrac{3}{4} \cdot \dfrac{(r + 6)(r - 7)}{(r + 6)(r + 8)}$ Assuming $r \neq -6$ , we can cancel the $r + 6$ $ z = - \dfrac{3}{4} \cdot \dfrac{r - 7}{r + 8}$ Therefore: $ z = \dfrac{ -3(r - 7)}{ 4(r + 8)}$, $r \neq -6$